Physics and Mathematics
Algebraic Limits By Using Standard Results
1. Concept Overview
In many algebraic limit problems, direct substitution leads to the indeterminate form [0/0].
Instead of factorisation or rationalisation every time, we can use standard limit results to evaluate such limits quickly and correctly.
Standard results are pre-proved limits that help us evaluate limits directly.
2. Why Standard Results Are Needed
Consider the limit:
[\lim_{x \to a} \dfrac{x^{n} – a^{n}}{x – a}]
If we substitute [x = a]:
Numerator → [a^{n} − a^{n} = 0]
Denominator → [a − a = 0]
So the limit becomes [0/0], which is indeterminate.
This is exactly where standard results are applied.
3. Important Standard Algebraic Limit
Standard Result
[\lim_{x \to a} \dfrac{x^{n} – a^{n}}{x – a} = n a^{n-1}]
Conditions
- [n] is a positive integer
- [a ≠ 0]
4. Meaning of This Standard Result
This result tells us:
When a power expression like [x^{n}] approaches [a^{n}], the ratio of their difference to [x − a] approaches the derivative-like value [n a^{n-1}].
Even though students don’t need calculus here, this result naturally arises from algebraic factorisation.
5. Verification Using Simple Algebra (Conceptual)
For small values of [n]:
Example: [n = 2]
[x^{2} − a^{2} = (x − a)(x + a)]
So,
[\dfrac{x^{2} − a^{2}}{x − a} ][= x + a]
Now substitute [x = a]:
[a + a = 2a]
Which matches:
[n a^{n-1} = 2a^{1} = 2a]
✔ Verified.
6. Related Standard Results (Used Frequently)
- [\lim_{x \to a} \dfrac{x – a}{x – a} = 1]
- [\lim_{x \to 0} \dfrac{x^{n}}{x} = 0] for [n > 1]
- [\lim_{x \to a} \dfrac{x^{2} – a^{2}}{x – a} = 2a]
- [\lim_{x \to a} \dfrac{x^{3} – a^{3}}{x – a} = 3a^{2}]
(All are special cases of the main standard result.)
7. When to Use This Standard Result
Use this result only when:
✔ Expression is of the form [x^{n} − a^{n}]
✔ Denominator is exactly [x − a]
✔ Direct substitution gives [0/0]
8. Key Features to Remember
- Applies only to algebraic expressions
- Avoids lengthy factorisation
- Extremely useful in exams
- Must check exact form before applying
9. Conceptual Questions with Detailed Solutions
1. Why does the limit [\lim_{x \to a} \dfrac{x^{n} – a^{n}}{x – a}] become indeterminate by direct substitution?
When we substitute [x = a], the numerator becomes [a^{n} − a^{n} = 0] and the denominator becomes [a − a = 0].
The form [0/0] does not represent a definite value. It only tells us that both numerator and denominator approach zero, but their rate of approach may be different. Hence, further analysis using standard results is required.
2. Why can’t we simply cancel [x − a] from the denominator without using a standard result?
Direct cancellation is mathematically unsafe because at [x = a], the factor [x − a] becomes zero.
Standard results are derived by simplifying the expression before substitution, ensuring we never divide by zero. This preserves mathematical correctness.
3. How does the standard result avoid lengthy factorisation in exams?
Factorising [x^{n} − a^{n}] becomes lengthy for large values of [n].
The standard result directly gives the final value [n a^{n−1}], saving time and reducing algebraic mistakes—very important in board and competitive exams.
4. Why must the denominator be exactly [x − a] to apply the standard result?
The result is derived assuming the denominator approaches zero at the same rate as the numerator.
If the denominator is [x − b] where [b ≠ a], this balance is lost and the result becomes invalid.
5. Why is it compulsory to check the form before applying the standard result?
Students often memorise the formula and apply it blindly.
However, if the limit does not give [0/0] after substitution, the standard result is not required and may lead to a wrong answer.
6. Can this standard result be applied when [x → 0]?
Yes. In this case, [a = 0].
The result becomes:
[\lim_{x \to 0} \dfrac{x^{n} – 0}{x – 0} = n × 0^{n−1}]
This is valid only when [n > 1].
7. Why is this topic important before learning continuity?
Continuity requires checking whether:
[\lim_{x \to a} f(x) = f(a)]
Standard results help evaluate limits quickly, making continuity problems easier and less error-prone.
8. Why do examiners prefer students to mention the standard result explicitly?
Writing the standard result shows conceptual clarity and logical reasoning.
It also fetches method marks, even if a minor arithmetic mistake occurs later.
9. Why does the power [n] have to be a positive integer?
The algebraic derivation relies on factorisation patterns that are valid only for positive integers.
Fractional or negative powers require different techniques.
10. How does this result connect algebra and calculus concepts?
Though taught algebraically, the result represents the slope of the curve [y = x^{n}] at [x = a].
This prepares students mentally for differentiation in calculus.
11. Why should beginners not skip the substitution step?
Skipping substitution hides the indeterminate nature of the limit.
Substitution confirms why we are allowed to use the standard result.
12. What conceptual mistake do beginners commonly make in this topic?
They apply the formula even when the denominator is not [x − a] or when the expression is already determinate.
13. Why is [0/0] called an indeterminate form?
Because it does not uniquely determine the value of the limit. Different functions producing [0/0] can approach different values.
14. How does this topic improve exam confidence?
It provides a direct, reliable method for a large class of questions, reducing fear and uncertainty during exams.
15. When should a student avoid using this standard result?
When:
The denominator is not [x − a]
The power is not an integer
The limit is not indeterminate
10. FAQs / Common Misconceptions
1. “If the limit gives [0/0], the answer must be zero.”
Wrong.
[0/0] gives no information. The actual value depends on how numerator and denominator approach zero.
2. “Standard results are shortcuts, not real mathematics.”
Incorrect.
They are rigorously derived results, not tricks.
3. “I can use the standard result even if the denominator is different.”
False.
The denominator must be exactly [x − a].
4. “Substitution is optional.”
Wrong.
Substitution is essential to identify indeterminacy.
5. “Factorisation and standard results give different answers.”
Incorrect.
Both lead to the same value.
6. “This result works for square roots.”
No.
Square roots require rationalisation.
7. “The power [n] can be fractional.”
Incorrect.
Only positive integers are allowed.
8. “I don’t need to mention the formula in exams.”
Wrong.
Mentioning it improves presentation and marks.
9. “Limits and continuity are unrelated topics.”
Incorrect.
Limits are the foundation of continuity.
10. “Memorising the result is enough.”
False.
Understanding when and why to use it is more important.
11. Practice Questions with Step-by-Step Solutions
Question 1. Evaluate:
[\lim_{x \to 2} \dfrac{x^{2} − 4}{x − 2}]
Step-by-Step Solution:
Substitute [x = 2] → [0/0]
Apply standard result with [n = 2, a = 2]
Value = [2 × 2 = 4]
Final Answer: [4]
Question 2. Evaluate:
[\lim_{x \to 3} \dfrac{x^{3} − 27}{x − 3}]
Step-by-Step Solution:
[0/0] obtained
Use standard result
[n = 3, a = 3]
Value = [3 × 3^{2} = 27]
Final Answer: [27]
Question 3. Evaluate:
[\lim_{x \to 1} \dfrac{x^{5} − 1}{x − 1}]
Step-by-Step Solution:
Step 1: Substitute [x = 1] to check the form
Numerator:
[1^{5} − 1 = 1 − 1 = 0]
Denominator:
[1 − 1 = 0]
So, the limit is of the indeterminate form [0/0].
Step 2: Compare with the standard result
We know the standard result:
[\lim_{x \to a} \dfrac{x^{n} − a^{n}}{x − a} = n a^{n−1}]
Here:
[n = 5]
[a = 1]
Step 3: Apply the standard result
Value of the limit:
[5 × 1^{4} = 5]
Final Answer:
[\lim_{x \to 1} \dfrac{x^{5} − 1}{x − 1} = 5]
Question 4. Evaluate:
[\lim_{x \to 4} \dfrac{x^{2} − 16}{x − 4}]
Step-by-Step Solution:
Step 1: Direct substitution
Numerator:
[4^{2} − 16 = 16 − 16 = 0]
Denominator:
[4 − 4 = 0]
So, the limit is [0/0], an indeterminate form.
Step 2: Identify the standard form
Compare with:
[\lim_{x \to a} \dfrac{x^{n} − a^{n}}{x − a}]
Here:
[n = 2]
[a = 4]
Step 3: Apply the standard result
Value of the limit:
[2 × 4^{1} = 8]
Final Answer:
[\lim_{x \to 4} \dfrac{x^{2} − 16}{x − 4} = 8]
Question 5. Evaluate:
[\lim_{x \to 2} \dfrac{x^{4} − 16}{x − 2}]
Step-by-Step Solution:
Step 1: Substitute [x = 2]
Numerator:
[2^{4} − 16 = 16 − 16 = 0]
Denominator:
[2 − 2 = 0]
So, the limit is [0/0].
Step 2: Match with standard result
Here:
Power of [x] is [n = 4]
Approaching value is [a = 2]
Step 3: Apply the standard formula
Limit value:
[4 × 2^{3} = 4 × 8 = 32]
Final Answer:
[\lim_{x \to 2} \dfrac{x^{4} − 16}{x − 2} = 32]
Question 6. Evaluate:
[\lim_{x \to a} \dfrac{x^{3} − a^{3}}{x − a}]
Step-by-Step Solution:
Step 1: Substitute [x = a]
Numerator:
[a^{3} − a^{3} = 0]
Denominator:
[a − a = 0]
So the form is [0/0].
Step 2: Identify the standard result
Here:
[n = 3]
[a = a]
Step 3: Apply the formula
Value of the limit:
[3a^{2}]
Final Answer:
[\lim_{x \to a} \dfrac{x^{3} − a^{3}}{x − a} = 3a^{2}]
Question 7. Evaluate:
[\lim_{x \to 1} \dfrac{x^{6} − 1}{x − 1}]
Step-by-Step Solution:
Step 1: Substitute [x = 1]
Numerator:
[1^{6} − 1 = 0]
Denominator:
[1 − 1 = 0]
Indeterminate form [0/0].
Step 2: Compare with standard form
Here:
[n = 6]
[a = 1]
Step 3: Apply standard result
Limit value:
[6 × 1^{5} = 6]
Final Answer:
[\lim_{x \to 1} \dfrac{x^{6} − 1}{x − 1} = 6]
Question 8. Evaluate:
[\lim_{x \to 5} \dfrac{x^{2} − 25}{x − 5}]
Step-by-Step Solution:
Step 1: Substitute [x = 5]
Numerator:
[25 − 25 = 0]
Denominator:
[5 − 5 = 0]
So the form is [0/0].
Step 2: Identify parameters
Here:
[n = 2]
[a = 5]
Step 3: Apply standard result
Limit value:
[2 × 5 = 10]
Final Answer:
[\lim_{x \to 5} \dfrac{x^{2} − 25}{x − 5} = 10]
Question 9. Evaluate:
[\lim_{x \to a} \dfrac{x^{n} − a^{n}}{x − a}]
Step-by-Step Solution:
Step 1: Substitute [x = a]
Numerator → [0]
Denominator → [0]
So the form is [0/0].
Step 2: Recognise the standard result
This expression exactly matches the standard algebraic limit.
Step 3: Write the value
Limit value:
[n a^{n−1}]
Final Answer:
[n a^{n−1}]
Question 10. Evaluate:
[\lim_{x \to 0} \dfrac{x^{3} − 0}{x}]
Step-by-Step Solution:
Step 1: Simplify the expression
Numerator:
[x^{3} − 0 = x^{3}]
So the limit becomes:
[\lim_{x \to 0} \dfrac{x^{3}}{x}]
Step 2: Simplify algebraically
[\dfrac{x^{3}}{x} = x^{2}]
Step 3: Substitute [x = 0]
[0^{2} = 0]
Final Answer:
[\lim_{x \to 0} \dfrac{x^{3} − 0}{x} = 0]
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